445_Physics ProblemsTechnical Physics - Chapter 15 P15.21(a...

Chapter 15447P15.21(a)EkA=122, so if ′ =AA2, ′ =′==Ek AkAE12122422a fafTherefore Eincreases by factor of 4.(b)vkmAmax=, so if Ais doubled, vmaxis doubled.(c)akmAmax=, so if Ais doubled, amaxalso doubles.(d)Tmk=2πis independent of A, so the period is unchanged.*P15.22(a)yyvta tfiyiy=++122−=++−=⋅=1100129 8229 81 502mm sm sms22...ejtt(b)Take the initial point where she steps off the bridge and the final point at the bottom of hermotion.KUUKUUmgykxkkgsigsf++=++++=++==ej ejaf000012659 8122573 422kg ms36 mmN m2..(c)The spring extension at equilibrium is xFk===658 68kg 9.8 m s73.4 N mm2., so this point is118 6819 7+=..mm below the bridgeand the amplitude of her oscillation is

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